Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{n^2 - 10n + 21}{5n + 50} \times \dfrac{n + 10}{-8n + 24} $
First factor the quadratic. $a = \dfrac{(n - 3)(n - 7)}{5n + 50} \times \dfrac{n + 10}{-8n + 24} $ Then factor out any other terms. $a = \dfrac{(n - 3)(n - 7)}{5(n + 10)} \times \dfrac{n + 10}{-8(n - 3)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (n - 3)(n - 7) \times (n + 10) } { 5(n + 10) \times -8(n - 3) } $ $a = \dfrac{ (n - 3)(n - 7)(n + 10)}{ -40(n + 10)(n - 3)} $ Notice that $(n + 10)$ and $(n - 3)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(n - 3)}(n - 7)(n + 10)}{ -40(n + 10)\cancel{(n - 3)}} $ We are dividing by $n - 3$ , so $n - 3 \neq 0$ Therefore, $n \neq 3$ $a = \dfrac{ \cancel{(n - 3)}(n - 7)\cancel{(n + 10)}}{ -40\cancel{(n + 10)}\cancel{(n - 3)}} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $a = \dfrac{n - 7}{-40} $ $a = \dfrac{-(n - 7)}{40} ; \space n \neq 3 ; \space n \neq -10 $